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Which compound is formed when excess of \(\mathrm{KCN}\) is added to an aqueous solution of copper sulphate? (a) \(\mathrm{Cu}(\mathrm{CN})_{2}\) (b) \(\mathrm{K}_{2}\left[\mathrm{Cu}(\mathrm{CN})_{6}\right]\) (c) \(\mathrm{K}\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]\) (d) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\)

Short Answer

Expert verified
(d) \( \mathrm{K}_{3}[\mathrm{Cu(CN)}_{4}] \)

Step by step solution

01

Identify the Reactants

The reactants in the problem are copper sulfate (\( \mathrm{CuSO}_4 \)) and potassium cyanide (\( \mathrm{KCN} \)). We are adding an excess of \( \mathrm{KCN} \) to \( \mathrm{CuSO}_4 \).
02

Initial Reaction Formation

\( \mathrm{KCN} \) is a source of \( \mathrm{CN}^- \) ions which can initially react with \( \mathrm{CuSO}_4 \) to precipitate \( \mathrm{Cu(CN)}_{2} \). However, \( \mathrm{Cu(CN)}_{2} \) is unstable and decomposes into \( \mathrm{CuCN} \) and \( \mathrm{C}_2\mathrm{N}_2 \).
03

Complexation with Excess CN^-

The \( \mathrm{CuCN} \) forms a complex with excess \( \mathrm{CN}^- \) ions in the solution, leading to the formation of a complex ion, \([\mathrm{Cu(CN)_4}]^{3-}\). This is because cyanide is a good ligand and forms stable complexes with copper.
04

Final Compound Formation

The formed complex \([\mathrm{Cu(CN)_4}]^{3-}\) binds with potassium ions from the \( \mathrm{KCN} \) solution, resulting in the formation of \( \mathrm{K}_3[\mathrm{Cu(CN)_4}] \), which can be identified as one of the options provided.
05

Select the Correct Option

Comparing arguments from our analysis, the compound that matches the chemical formation is \( \mathrm{K}_{3}[\mathrm{Cu(CN)}_{4}] \). This corresponds to option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Ion Formation
In the realm of coordination chemistry, complex ions are formed when central metal ions bond with molecules or ions, called ligands. This process occurs because these metal ions typically have empty d-orbitals capable of accepting electron pairs from the ligands. A great example is the reaction between copper sulfate (\( \mathrm{CuSO}_4 \)) and potassium cyanide (\( \mathrm{KCN} \)), where the cyanide ions (\( \mathrm{CN}^- \)) act as ligands.
  • Initially, copper ions form a complex with available cyanide ions to create \( \mathrm{Cu(CN)_2} \).
  • However, this initially formed complex is unstable and requires further interaction with additional cyanide ions to reach stability.
Thus, in the presence of excess cyanide ions from the excess \( \mathrm{KCN} \), the copper ions are enclosed by four cyanide ligands forming a stable complex entity \([\mathrm{Cu(CN)_4}]^{3-}\). This advanced bonding scenario involves coordination bonds where cyanides use their lone pairs to bond with copper, a typical feature of complex ion formation.
Ligand Exchange
Ligand exchange is a crucial concept in coordination chemistry as it relates to the swapping of ligands around a central metal ion. This phenomenon allows for the formation and transformation of complex ions as observed in our exercise. When excess \( \mathrm{KCN} \) is added to a \( \mathrm{CuSO}_4 \) solution, the cyanide ions (\( \mathrm{CN}^- \)) initially bond with copper, forming a slightly unstable compound, \( \mathrm{Cu(CN)_2} \). The magic happens when more \( \mathrm{CN}^- \) ions in the solution facilitate a ligand exchange:
  • The unstable \( \mathrm{Cu(CN)_2} \) decomposes into \( \mathrm{CuCN} \) and \( \mathrm{C}_2\mathrm{N}_2 \)
  • Subsequently, the \( \mathrm{CuCN} \) undergoes ligand exchange with excess \( \mathrm{CN}^- \), forming the complex ion \([\mathrm{Cu(CN)_4}]^{3-}\)
This exchange ensures a more stable structure, and the ability to achieve such stability is integral for the further application of coordination compounds in various fields like catalysis and medicine.
Stability of Complexes
The stability of complex ions is a cornerstone concept in coordination chemistry. Stability is significantly influenced by the nature of the ligands and their ability to donate electron pairs. When dealing with copper ions, cyanide \( \mathrm{CN}^- \) ions are exceptional at stabilizing the complex.
  • Cyanide is a strong field ligand meaning it can donate electrons effectively, stabilizing the positive charge of the metal ion.
  • This results in the formation of stronger coordination bonds, due to the Ï%s pairing of the electron clouds between the ligand and the metal ion as seen in \([\mathrm{Cu(CN)_4}]^{3-}\)
Such a stable complex formation minimizes energy and maximizes stability, rendering the ion less reactive towards other possible ligand exchanges. Stable complexes, such as \( \mathrm{K}_3[\mathrm{Cu(CN)_4}] \), therefore tend to be crucial in applications that require long-term chemical stability. They find usage in various industries including electroplating and medical imaging, where resilience of the compound is essential.

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Most popular questions from this chapter

The soldiers of Napolean's army, while on Alps during freezing winter, suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to (a) an interaction with nitrogen of the air at very low temperatures (b) a change in the crystalline structure of tin (c) an interaction with water vapour contained in the humid air (d) a change in the partial pressure of oxygen in the air.

Excess of \(\mathrm{KI}\) react with \(\mathrm{CuSO}_{4}\) solution and then \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution is added to it. Which of the following statements is incorrect for this reaction? (a) \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) is formed (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is oxidized (c) \(\mathrm{CuI}_{2}\) is formed (d) evolved \(\mathrm{I}_{2}\) is reduced

The transition elements with some exceptions can show a large number of oxidation states. The various oxidation states are related to the electronic configuration of their atoms. The variable oxidation states of a transition metal is due to the involvement of \((\mathrm{n}-1) \mathrm{d}\) and outer \(\mathrm{ns}\)-electrons. For the first five elements of 3 d-transition series. The minimum oxidation state is equal to the number of electrons in 4s shell and the maximum oxidation state is equal to the sum of \(4 \mathrm{~s}\) and \(3 \mathrm{~d}\)-electrons. The relative stability of various oxidation state of a given element can be explained on the basis of stability of \(\mathrm{d}^{0}, \mathrm{~d}^{5}\) and \(\mathrm{d}^{10}\) configurations. In 3 d-series, the maximum oxidation state is shown by (a) \(\mathrm{Fe}(26)\) (b) Mn (atomic no.: 25 ) (c) \(\mathrm{Cr}(24)\) (d) Sc (atomic no: 21 )

Transition metals despite having higher values of standard reduction potentials, are poor reducing agents. This is due to (a) low heat of hydration (b) high ionization energies (c) low ionization energies (d) high enthalpy of vapourization

\(\mathrm{CuSO}_{4}\) decolourizes on addition of KCN, the product is (a) \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}\) (b) \(\mathrm{Cu}^{2+}\) gets reduced to form \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) (c) \(\mathrm{Cu}(\mathrm{CN})_{2}\) (d) \(\mathrm{CuCN}\)

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